A Wideband Code Division Multiple Access (WCDMA) system provides that a user equipment sends multiple control channels and data channels. These channels are made orthogonal to each other through the use of channelization codes. Therefore, these channels may be transmitted simultaneously, and a receiver can separate them using the different channelization codes.
However, in a multipath scenario, any two channels from different fingers (multipaths) are no longer orthogonal because their arrival timing at the receiver is mismatched due to the latency between the two fingers. Therefore, while a receiver separates a current channel from other channels in the same finger using the associated channelization code, the channels from other fingers may leak into the current channel. This leakage is called cross-finger interference. The energy of the cross-finger interference is the original energy in the other finger reduced by a despreading factor determined by the symbol size of the channelization code in the current finger of the receiver.
FIG. 1 illustrates a channel model for a Wideband Code Division Multiple Access (WCDMA) system. Here, a user equipment (UE) has multiple concurrent channels, such as Sc(t) for a control channel signal (e.g., a Dedicated Physical Control Channel or DPCCH in WCDMA) and Sd(t) for a data channel signal (e.g., E-DCH Dedicated Physical Data Control Channel or E-DPDCH in WCDMA). While the UE may send many concurrent channels, only two channels are shown for the sake of brevity and simplicity of disclosure. In particular, the UE applies respective orthogonal codes at spreaders 10, 12 for each signal, and a transmitter 14 transmits the multiple signals over a medium such as an air interface. Noise n(t) is an additive channel noise added by transmission over the medium. A base station or NodeB therefore receives the following waveform at its receiver 20:x(t)=Sc(t)+Sd(t)+n(t)
In this example, if the control signal and the data signal come from different fingers (or multi-paths), then power of one signal may leak into the power of another signal causing cross-finger interference. This interference significantly impacts receiver performance, particularly when determining a signal-to-noise ratio (SIR) and the operations such as uplink power control which rely on the SIR determination.
To better understand this impact, uplink power control will be briefly discussed. As is well-known, during uplink power control according to WCDMA, a NodeB and UE establish a power control loop only based on the DPCCH. The NodeB measures the SIR (DPCCH Sc(t) power to n(t) noise power ratio) and compares the measured SIR to a target SIR. The NodeB instructs the UE to adjust the uplink DPCCH power level to meet the desired SIR range and a desired bit error rate (BER) performance received, for example, from a radio network controller (RNC). After this power control closed loop is established, the UE will generate a DPCCH uplink transmission gain factor βc for the DPCCH channel. The UE uses the gain factor βc as a base to calculate the transmission power for the other uplink channels. For example, for the E-DPDCH channel, a desired power ratio is defined by TPR=(E-DPDCH power)/(DPCCH power). The E-DPDCH transmission power gain factor βed is then determined according to the following expression:βed2=TPR*βc2 
In this approach, the closed loop power control should not include E-DPDCH signal power. Any interference leaked from the E-DPDCH into the DPCCH will cause the DPCCH SIR measurement to have an offset and result in misleading the power control loop. If the cross-finger interference is too strong, the problem is that the power control loop diverges and can not set a correct power gain factor for each channel.
The cross-finger interference could be very strong and significantly impact system performance. For example, in WCDMA system, E-DPDCH power may be 100s of times stronger than the DPCCH. For the DPCCH, the cross-finger interference from the E-DPDCH could be even stronger than DPCCH itself. For additive channel noise power estimation, the cross-finger interference could cause a huge offset for the noise power estimation. This cross-finger interference may cause a NodeB to miss-measure the SIR and lead to power diverge in uplink power control loop.
Furthermore, existing techniques for estimating SIR rely on a correlation function as shown by the expressions below. First, assume the output symbol at the receiver for the control signal Sc is:y(t)=Sc(t)+n(t), where t is a symbol index.
At the receiver, the control signal power E[Sc2] is determined based on correlation function as follows:
                                          E            ⁡                          [                              Sc                2                            ]                                ∼                ⁢                =                              y            ⁡                          (              t              )                                *                      y            ⁡                          (                              t                -                1                            )                                                                      ⁢                  =                                    [                                                Sc                  ⁡                                      (                    t                    )                                                  +                                  n                  ⁡                                      (                    t                    )                                                              ]                        *                          [                                                Sc                  ⁡                                      (                                          t                      -                      1                                        )                                                  +                                  n                  ⁡                                      (                                          t                      -                      1                                        )                                                              ]                                                                      ⁢                  =                                                    Sc                ⁡                                  (                  t                  )                                            ⁢                              Sc                ⁡                                  (                                      t                    -                    1                                    )                                                      +                          {                                                                    Sc                    ⁡                                          (                      t                      )                                                        ⁢                                      n                    ⁡                                          (                                              t                        -                        1                                            )                                                                      +                                                      Sc                    ⁡                                          (                                              t                        -                        1                                            )                                                        ⁢                                      n                    ⁡                                          (                      t                      )                                                                      +                                                      n                    ⁡                                          (                      t                      )                                                        ⁢                                      n                    ⁡                                          (                                              t                        -                        1                                            )                                                                                  }                                                                      ⁢                  =                                                    Sc                ⁡                                  (                  t                  )                                            ⁢                              Sc                ⁡                                  (                                      t                    -                    1                                    )                                                      +                                          o                c                            ⁡                              (                                  Sc                  ,                  n                                )                                                                                    ⁢                  ->                      Sc            2                              where oc(Sc, n) is called a correlation remainder.
The receiver determines the noise power E[n2] according to the following expression:
                                          E            ⁡                          [                              n                2                            ]                                ∼                ⁢                =                                            y              ⁡                              (                t                )                                      2                    -                                    y              ⁡                              (                t                )                                      ⁢                          y              ⁡                              (                                  t                  -                  1                                )                                                                                    ⁢                  =                                                    [                                                      Sc                    ⁡                                          (                      t                      )                                                        +                                      n                    ⁡                                          (                      t                      )                                                                      ]                            2                        -                          [                                                Sc                  2                                +                                                      o                    c                                    ⁡                                      (                                          Sc                      ,                      n                                        )                                                              ]                                                                      ⁢                  =                                    [                                                                    Sc                    ⁡                                          (                      t                      )                                                        2                                +                                                      n                    ⁡                                          (                      t                      )                                                        2                                +                                  2                  ⁢                                      Sc                    ⁡                                          (                      t                      )                                                        ⁢                                      n                    ⁡                                          (                      t                      )                                                                                  ]                        -                          [                                                Sc                  2                                +                                                      o                    c                                    ⁡                                      (                                          Sc                      ,                      n                                        )                                                              ]                                                                      ⁢                  =                                                    n                ⁡                                  (                  t                  )                                            2                        +                                          o                n                            ⁡                              (                                  Sc                  ,                  n                                )                                                                                    ⁢                  ->                      n            2                              where on(Sc, n) is also called correlation remainder.
These measurements include both signal power and cross-finger interference power, which may be represented as:Ek(Sc(t)2=EkSc+Ecross2—k, where k represents a finger indexEk[n(t)2]=Ekn+Ecross1—k where Ek(Sc(t)2) is the received control signal power, EkSc represents the portion of the received control signal power due to the sent control signal, Ecross2_k represents the portion of the control signal power due to the cross-finger interference, Ek[n(t)2] is the received noise power, Ekn represents the portion of the received noise power due to the additive channel noise, and Ecross1_k represents the portion of the received noise power due to the cross-finger interference. In view of the above, the SIR may be expressed as:
                    SIR        =                              ∑            k            fingers                    ⁢                                          ⁢                                                    E                k                            ⁡                              [                                                      Sc                    ⁡                                          (                      t                      )                                                        2                                ]                                                                    E                k                            ⁡                              [                                                      n                    ⁡                                          (                      t                      )                                                        2                                ]                                                                            =                              ∑            k            fingers                    ⁢                                          ⁢                                                                      E                  k                                ⁢                Sc                            +                              E                                  cross                  ⁢                  2                  ⁢                  _k                                                                                                      E                  k                                ⁢                n                            +                              E                                  cross                  ⁢                  1                  ⁢                  _k                                                                        
Both the signal power and noise power estimation output contain cross-finger interference since the cross-finger interference exists in the input of the correlation function, or output of DPCCH despreader. In a High Speed Uplink Packet Access (HSUPA) environment, the cross-finger interference could be much stronger than noise power:Ecross1—k>>Ek[n(t)2]This causes SIR measurements to have a huge random offset, and significantly downgrades throughput.